Find $\dfrac{d}{dx}\left[2^{\cot(5x)}\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $5\ln(2)\cdot 2^{ \cot(5x)}\cdot\sin^2(5x)$ (Choice B) B $\dfrac{-5\cdot 2^{ \cot(5x)}}{\sin^2(5x)}$ (Choice C) C $5\ln(2)\cdot 2^{ \cot(5x)}$ (Choice D) D $\dfrac{-5\ln(2)\cdot 2^{ \cot(5x)}}{\sin^2(5x)}$
Explanation: $2^{\cot(5x)}$ is a composition of three functions! Let... $u(x)=2^x$ $v(x)=\cot(x)$ $w(x)=5x$... then $2^{\cot(5x)}=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $\dfrac{d}{dx}\left[2^{\cot(5x)}\right]$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\ln(2)\cdot 2^x$ $v'(x)=-\dfrac{1}{\sin^2(x)}$ $w'(x)=5$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={\ln(2)\cdot 2^{\cot(5x)}}\cdot{ -\dfrac{1}{\sin^2(5x)}}\cdot{5} \\\\ &=\dfrac{-5\ln(2)\cdot 2^{{\cot(5x)}}}{\sin^2(5x)}\end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[2^{\cot(5x)}\right]=\dfrac{-5\ln(2)\cdot 2^{{\cot(5x)}}}{\sin^2(5x)}$.